Question

The indicated function y1(x) is a solution of the given differential equation. Use reduction of order or formula (5) in Section 4.2, y2 = y1(x) e−∫P(x) dx y 2 1 (x) dx (5) as instructed, to find a second solution y2(x). y'' + 2y' + y = 0; y1 = xe−x

Answer 1

First confirm that $y_1=xe^{-x}$ is a solution to the ODE,

$y''+2y'+y=0$

We have

${y_1}'=e^{-x}-xe^{-x}=(1-x)e^{-x}$

${y_1}''=-e^{-x}-(1-x)e^{-x}=(-2+x)e^{-x}$

Substituting into the ODE gives

$(-2+x)e^{-x}+2(1-x)e^{-x}+xe^{-x}=0$

Suppose $y_2(x)=v(x)y_1(x)$ is another solution to this ODE. Then

${y_2}'=v'y_1+v{y_1}'$

${y_2}''=v''y_1+2v'{y_1}'+v{y_1}''$

and substituting these into the ODE yields

$(v''y_1+2v'{y_1}'+v{y_1}'')+2(v'y_1+v{y_1}')+vy_1=0$

$xe^{-x}v''+2e^{-x}v'=0$

$xv''+2v'=0$

Let $w(x)=v'(x)$. Then the remaining ODE is linear in $w$:

$xw'+2w=0$

Multiply both sides by the integrating factor, $x$, and condense the left hand side as a derivative of a product:

$x^2w'+2xw=(x^2w)'=0$

Integrate both sides with respect to $x$ and solve for $w$:

$x^2w=C_1\implies w=C_1x^{-2}$

Back-substitute and integrate both sides with respect to $x$ to solve for $v$:

$v'=C_1x^{-2}\implies v=-C_1x^{-1}+C_2$

Back-substitute again to solve for $y_2$:

$\dfrac{y_2}{y_1}=C_2-\dfrac{C_1}x$

$\implies y_2=C_2xe^{-x}-C_1e^{-x}$

$y_1$ already captures the solution $xe^{-x}$, so the remaining one is

$\boxed{y_2=e^{-x}}$